DOCSLIB.ORG

Magnetic Dipole Field

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Magnetic Dipole Field Fundamentals of Magnetism Albrecht Jander Oregon State University Part I: H, M, B, χ, µ Part II: M(H) Magnetic Field from Current in a Wire H dl I r R c dl Ampere’s Circuital Law: Biot-Savart Law: × Jean-Baptiste Biot = = (1774 –1862) c 퐼4푑푙⃗ 푅� c � 퐻 ∙ 푑푙⃗ 퐼 푑퐻 2 휋 푅 = [A/m] 퐼 André-Marie Ampère 2휋휋 Félix Savart (1775–1836) 퐻 (1791 – 1841) Magnetic Field from Current Loop H r I Use Biot-Savart Law: In the center: × = = [A/m] 퐼4푑푙⃗ 푅� 푑퐻 2 퐼 휋 푅 퐻 2휋 Magnetic Field in a Long Solenoid H I JS I N turns L Inside, field is uniform with: = [A/m] = [A/m] 푁퐼 푆 퐻 퐿 퐻 퐽 Magnetic Field from “Small” Loop (Dipole) H r I Magnetic Field from Small Loop (Dipole) H R θ Compare to Electric dipole: For R >> r = 2 cos + sin ( ) [A/m] 4 퐼퐼 퐻 3 풓� 휃 휽� 휃 Magnetic (dipole)휋 moment:푅 = [Am2] 푚 퐼퐼⃗ Magnetic Field of a Dipole H R θ [A·m] For R >> r 푞푚 = 2 cos + sin ( ) [A/m] 4 푞푚푑 Magnetic퐻 (dipole) moment:3 푟̂ 휃 휃� 휃 휋푅 = [Am2] 푚 푞푚푑⃗ Magnetic Poles and Dipoles Electric charge Magnetic charge, or “pole” + q + qm = 4 푞 [A·m] 퐷 2 휋푅 Magnetic field from a magnetic “monopole” 푞푚 = [A/m] 4 푞푚 퐻 2 휋푅 Magnetic Field of a Dipole + qm d ̶ qm [A·m] 푞푚 Magnetic Field of a Dipole H R θ [A·m] For R >> r 푞푚 = 2 cos + sin ( ) [A/m] 4 푞푚푑 Magnetic퐻 (dipole) moment:3 푟̂ 휃 휃� 휃 휋푅 = [Am2] 푚 푞푚푑⃗ Orbital Magnetic Moment •Electrons orbiting a nucleus are like a circulating current producing e- a magnetic field. •For an electron with charge qe orbiting at a radius R with frequency f, the Orbital Magnetic Moment is 2 2 m = IA = −qe f πR [Am ] • It also has an Orbital Angular Momentum L = mevR = me 2πRf R • Note: m − q = e = gyromagnetic ratio L 2me Orbital Moment is Quantized Bohr model of the atom λdB •DeBroglie wavelength is: e- h R λdB = mev •Bohr Model: orbit must be integer number of wavelengths h h 2πR = NλdB = N = N mev me 2πRf • Thus the orbital magnetic moment is quantized: 2 h qe qe m = −qe f πR = N = N 2π 2me 2me • Magnetic moment restricted to multiples of Bohr Magneton qe −24 2 µB = = 9.2742×10 [Am ] Louis de Broglie (1892-1987) 2me Spin • Spin is a property of subatomic particles (just like charge or mass.) • A particle with spin has a magnetic dipole moment and angular momentum. Pauli and Bohr contemplate • Spin may be thought of as conceptually the “spin” of a tippy-top arising from a spinning sphere of charge. (However, note that neutrons also have spin but no charge!) Electron Spin • When measured in a particular direction, the measured angular momentum of an electron is L = ± z 2 • When measured in a particular direction, the measured magnetic moment of an electron is Compare: qe mz = sz = ±µB Spin Orbital me mz = ±µB mz = NµB • We say “spin up” and “spin down” m − q m − q = e = e • Note: m − q L me L 2me = e L me m − q g = 2 = g e g =1 L 2me Stern-Gerlach Experiment - 1922 Demonstrated that magnetic moment is quantized with ±µB Otto Stern (1888-1969) Walther Gerlach, Otto Stern (1922). "Das magnetische Moment des Walter Gerlach Silberatoms". Zeitschrift für Physik A Hadrons and Nuclei 9. (1889-1979) Stern-Gerlach Experiment Quantum Numbers of Electron Orbitals •Electrons bound to a nucleus move in orbits identified by quantum numbers (from solution of Schrödinger's equation): n 1… principal quantum number, identifies the “shell” l 0..n-1 angular momentum q. #, type of orbital, e.g. s,p,d,f ml -l..l magnetic quantum number ms +½ or -½ spin quantum number For example, the electron orbiting the Spectroscopic notation hydrogen nucleus (in the ground state) has: l=0 s l=1 p n=1, l=0, m =0, m =+½ l s l=2 d l=3 f In spectroscopic notation: 1s1 n l # of electrons in orbital 1s22s2 Spin and Orbital Magnetic Moment • Total orbital magnetic moment (sum over all electron orbitals) mtot _ orbital = µB ∑ ml • Total spin magnetic moment mtot _ spin = 2µB ∑ ms full orbitals: no net moment E.g. Iron: 1s22s22p63s23p63d64s2 How to fill remaining d orbitals (l=2) 6 electrons for 10 spots: ml=-2 ml=-1 ml=0 ml=+1 ml=+2 Hund’s rules: ms=+½ Maximize Σms ms=-½ Then maximize Σml m = 2µ m = 4µ tot _ orbital B tot _ spin B Magnetization, M Atomic magnetic moment: What is H here? [Am2] M Object푚 magnetic푎푎푣푚 moment: 2 . = [Am ] Magnetic moment per unit volume: Atomic푚표표표 density:∑ 푚푎푎표푚 . [1/m3] = [A/m] = 푁 푚표표표 푑 푀 ≝ 푚푎푎표푚푑 푣푣푙 푣푣푙 Equivalent Loop Currents Atomic magnetic moment: 퐼 What is H here? [Am2] = I M Object푚푎푎 magnetic푣푚 moment:퐼⃗ 2 . = [Am ] Magnetic moment per unit volume: Atomic푚표표표 density:∑ 푚푎푎표푚 . [1/m3] = [A/m] = 푁 푚표표표 푑 푀 ≝ 푚푎푎표푚푑 푣푣푙 푣푣푙 Equivalent Surface Current Atomic magnetic moment: 퐼 What is H here? = I [Am2] M 푛� [A/m] 푎푎푣푚 J Object푚 magnetic moment:퐼⃗ S 2 . = [Am ] Equivalent surface current: Atomic푚표표표 density:∑ 푚푎푎표푚 = × [A/m] [1/m3] = 푁 퐽⃗푆 푀 푛� 푑 푣푣푙 Magnetic Pole Model Atomic magnetic moment: What is H here? 2 = [Am ] M Object푚푎푎 magnetic푣푚 푞 푚moment:푑⃗ 2 . = [Am ] Magnetic moment per unit volume: Atomic푚표표표 density:∑ 푚푎푎표푚 [A/m] = [1/m3] 푚 푁 푀 ≝ 푑 푣푣푙 푣푣푙 Equivalent Surface Pole Density Surface charge density: [Am] = [m2] 푞푚 Atomic magnetic moment: 휌푚푚 2 푎푟푎푎 = [Am ] M Object푚푎푎 magnetic푣푚 푞 푚moment:푑⃗ 2 . = [Am ] Magnetic pole density: Atomic푚표표표 density:∑ 푚푎푎표푚 = [A/m] [1/m3] = 푁 휌푚푚 푀 ∙ 푛� 푑 푣푣푙 Example: NdFeB Cylinder 3 cm A 2 cm M JS 6 M =10 [A/m] 6 [A/m] JS =10 -6 [m3] [A] vol 14·10 I = 20000 -4 [m2] 2 A 7·10 m =≃ 14 [Am ] m = IA = 14 [Am2] ≃ Example: NdFeB Cylinder 3 cm + + ++ + + ρ + + ++ + + + ms 2 cm M 6 [A/m] 6 M =10 ρms =10 [A/m] -6 [m3] -4 [m2] vol 14·10 A 7·10 [Am] 2 q = 700 m =≃ 14 [Am ] m≃ -2 [m] d =2·10 [Am2] m = qmd = 14 Example: NdFeB Cylinder + + + +++ + + + + + + + M Result is the same external to magnetic material. Hint: we are “far” away from the dipoles!!! Example: NdFeB Cylinder + + + +++ + + + + + + + M Result is different inside the magnetic material. The Constitutive Relation = ( + ) 푩 흁ퟎ 푯 푴 B = Magnetic flux density [ Tesla ] H = Magnetic field, “Magnetizing force” [A/m] M = Magnetization [A/m] 2 µ0 = Magnetic constant, “Permeability of free space” [Tesla-m/A] [Henry/m] [N/A ] Note, in vacuum, M is zero: = 푩 흁ퟎ푯 What is µo ? µ0 comes from the SI definition of the Ampere: The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to 2×10−7 newton per meter of length. I 1 = = I 2 흁ퟎ푰ퟏ 푩 흁ퟎ푯 F = × ퟐ흅풓 B 푭 푰ퟐ푳 푩 Maxwell’s Equations (Magnetostatics) Differential form Integral form = = 훁 ∙ 푩 ퟎ � 푩 ∙ 풅풔 ퟎ × = = 훁 푯 푱⃗ � 푯 ∙ 풅풍⃗ 푰 With: = ( + ) ퟎ 푩 흁= 푯 ( 푴 + ) = = [A/m2] 훁 ∙ 푩 흁ퟎ 훁 ∙ 푯 훁 ∙ 푴 훁 ∙ 푯 −훁 ∙ 푴 흆풎 Magnetic charge density i.e. magnetic charge/volume Example: NdFeB Cylinder + + ++ + + + + ++ + + + M B H = ( + ) 푩 흁ퟎ 푯 푴 Boundary Conditions B1┴ B2┴ H1|| H2|| The perpendicular component of B The transverse component of H is is continuous across a boundary continuous across a boundary (unless there is a true current on the boundary) Demagnetizing Fields + + ++ + + + + ++ + + + M Hd 푯풅 ∝ 푴 = 푯풅 −푵푴 Demagnetizing Factor Demagnetizing Factors: Special Cases = + + = 푯풅 −푵푴 푵풙 푵풚 푵풛 ퟏ Sphere Thin Sheet Long Rod = = = = = ퟏ = 푵풙 푵풚 푵풛 푵풙 푵풚 ퟎ z ퟑ 풛 = = 푵 ퟏ x ퟏ 푵풙 푵=풚 y ퟐ 푵풛 ퟎ Demagnetizing Factors “Demag” fields in ellipsoids of revolution are uniform. = 푯풅 + −푵+푴 = 푵풙 푵풚 푵풛 ퟏ Example: Ba-Ferrite Sphere Barrium Ferrite: ~ [kA/m] 푴 ퟏퟏퟎ M H B = = = = ( + ) 풙 풚 풛 ퟏ 푵 푵 푵 푩 = 흁ퟎ(푯 푴[kA/m] ) = ퟑ ퟎ 풅 푩 = 흁. ퟏퟎퟎ [Tesla] 푯 =−푵푴 [kA/m] 푯풅 −ퟏퟎ 푩 ퟎ ퟏퟐퟏ Example: Ba-Ferrite Ellipsoid Barrium Ferrite: ~ [kA/m] 푴 ퟏퟏퟎ M H B = = . = ( + ) 푵풙 푵풚 ퟎ ퟒퟏ = ퟎ( . [kA/m] ) = 푩 흁 푯 푴 ퟎ 풅 푩 = 흁. ퟖퟐ ퟏ[Tesla] 푯 =−푵푴. [kA/m] 푯풅 −ퟏퟔ ퟏ 푩 ퟎ ퟏퟎퟒ Example: Ba-Ferrite Ellipsoid Barrium Ferrite: ~ [kA/m] 푴 ퟏퟏퟎ M H B = . = ( + ) 푵풛 ퟎ ퟏ = ퟎ( [kA/m] ) = 푩 흁 푯 푴 ퟎ 풅 푩 = 흁. ퟏퟑퟏ [Tesla] 푯 =−푵푴 [kA/m] 푯풅 −ퟏퟏ 푩 ퟎ ퟏퟔ Susceptibility and Permeability Ideal, linear soft magnetic material Ideal permanent magnet M M H H = Real, useful magnetic material Susceptibility [unitless] M 푴 흌푯 = ( + ) 푩 = 흁ퟎ(푯+ 흌)푯 H 푩 = 흁ퟎ ퟏ 흌 푯 Rel. Permeability 푩 흁ퟎ흁풓푯 [unitless] Example, Long Rod in a Long Solenoid I =1 A = 1000 = 999 N/L = 1000/m 휇휋 휒 = = 1 [kA/m] 푁퐼 = = = 퐻 퐿 999 [kA/m] 푁퐼 푀= 휒퐻 +휒 퐿 = (1000kA/m)= 1.25 [Tesla] or: 0 퐵 휇0 퐻 푀 휇 = = 1.25 [Tesla] Note: without the rod: = = 0.00125 [Tesla] 퐵 휇0휇휋퐻 퐵 휇0퐻 Example, Soft Magnetic Ball in a Field Hext M Hd B = = + = 푴 흌푯 흌 푯풆풙풆 푯풅 흌 푯풆풙풆 − 푵푴 = = < ! + + 흌 흌 푴 푯풆풙풆 흌풆풆풆 ퟑ‼ ퟏ 푵흌 ퟏ 푵흌 Example, Soft Magnetic Ball in a Field B = = + = 푴 흌푯 흌 푯풆풙풆 푯풅 흌 푯풆풙풆 − 푵푴 = = < ! + + 흌 흌 푴 푯풆풙풆 흌풆풆풆 ퟑ‼ ퟏ 푵흌 ퟏ 푵흌 Example, Thin Film in a Field Hext = = + = 푴 흌푯풊풊풔풊 흌 푯풆풙풆 푯풅 흌 푯풆풙풆 − 푴 = + 흌 푴 푯풆풙풆 ퟏ 흌 = = + ퟏ 푯풊풊풔풊 푯풆풙풆 푩풊풊풔풊 퐵Why?풆풙풆 ퟏ 흌 Torque and Zeeman Energy F + + + +++ + + Force: + + + + + = B M Torque: 퐹⃗ 푞푚퐵 = × F Energy: 풯 푚 퐵 = 퐸 −푚 ∙ 퐵 Self Energy and Shape Anisotropy M = Hd M 1 Hd 2 0 퐸 � 휇 푀 ∙ 퐻푑푣 Higher energy Lower energy 푣표표 “Shape Anisotropy” = ( ) E 2 Hard Hard 1 2 퐸 2휇0푀푚 푁푥 − 푁푌 푠푠푛 휃 Easy axis Easy Easy M θ 0 180 360 Hard axis Magnetic Circuits path length, l I = area,A N turns � 푯 ∙ 풅풍⃗ 푵푰 Assuming flux is uniform and contained inside permeable material: = = 흓 = 푩 흁ퟎ흁풓푯 푨 Magneto-motive force, “MMF”푵푰 흓흓 Reluctance, = 풍 Flux 흓 흁푨 Magnetic Circuits 흓ퟏ 흓ퟐ NI ퟑ 품품품 흓 흓품품품 흓 흓ퟒ 흓ퟏ Solve using electric circuit theory.
Load more

Recommended publications
  • Basic Magnetic Measurement Methods
    Basic magnetic measurement methods Magnetic measurements in nanoelectronics 1. Vibrating sample magnetometry and related methods 2. Magnetooptical methods 3. Other methods Introduction Magnetization is a quantity of interest in many measurements involving spintronic materials ● Biot-Savart law (1820) (Jean-Baptiste Biot (1774-1862), Félix Savart (1791-1841)) Magnetic field (the proper name is magnetic flux density [1]*) of a current carrying piece of conductor is given by: μ 0 I dl̂ ×⃗r − − ⃗ 7 1 - vacuum permeability d B= μ 0=4 π10 Hm 4 π ∣⃗r∣3 ● The unit of the magnetic flux density, Tesla (1 T=1 Wb/m2), as a derive unit of Si must be based on some measurement (force, magnetic resonance) *the alternative name is magnetic induction Introduction Magnetization is a quantity of interest in many measurements involving spintronic materials ● Biot-Savart law (1820) (Jean-Baptiste Biot (1774-1862), Félix Savart (1791-1841)) Magnetic field (the proper name is magnetic flux density [1]*) of a current carrying piece of conductor is given by: μ 0 I dl̂ ×⃗r − − ⃗ 7 1 - vacuum permeability d B= μ 0=4 π10 Hm 4 π ∣⃗r∣3 ● The Physikalisch-Technische Bundesanstalt (German national metrology institute) maintains a unit Tesla in form of coils with coil constant k (ratio of the magnetic flux density to the coil current) determined based on NMR measurements graphics from: http://www.ptb.de/cms/fileadmin/internet/fachabteilungen/abteilung_2/2.5_halbleiterphysik_und_magnetismus/2.51/realization.pdf *the alternative name is magnetic induction Introduction It
    [Show full text]
  • The Magnetic Moment of a Bar Magnet and the Horizontal Component of the Earth’S Magnetic Field
    260 16-1 EXPERIMENT 16 THE MAGNETIC MOMENT OF A BAR MAGNET AND THE HORIZONTAL COMPONENT OF THE EARTH’S MAGNETIC FIELD I. THEORY The purpose of this experiment is to measure the magnetic moment μ of a bar magnet and the horizontal component BE of the earth's magnetic field. Since there are two unknown quantities, μ and BE, we need two independent equations containing the two unknowns. We will carry out two separate procedures: The first procedure will yield the ratio of the two unknowns; the second will yield the product. We will then solve the two equations simultaneously. The pole strength of a bar magnet may be determined by measuring the force F exerted on one pole of the magnet by an external magnetic field B0. The pole strength is then defined by p = F/B0 Note the similarity between this equation and q = F/E for electric charges. In Experiment 3 we learned that the magnitude of the magnetic field, B, due to a single magnetic pole varies as the inverse square of the distance from the pole. k′ p B = r 2 in which k' is defined to be 10-7 N/A2. Consider a bar magnet with poles a distance 2x apart. Consider also a point P, located a distance r from the center of the magnet, along a straight line which passes from the center of the magnet through the North pole. Assume that r is much larger than x. The resultant magnetic field Bm at P due to the magnet is the vector sum of a field BN directed away from the North pole, and a field BS directed toward the South pole.
    [Show full text]
  • Magnetic Polarizability of a Short Right Circular Conducting Cylinder 1
    Journal of Research of the National Bureau of Standards-B. Mathematics and Mathematical Physics Vol. 64B, No.4, October- December 1960 Magnetic Polarizability of a Short Right Circular Conducting Cylinder 1 T. T. Taylor 2 (August 1, 1960) The magnetic polarizability tensor of a short righ t circular conducting cylinder is calcu­ lated in the principal axes system wit h a un iform quasi-static but nonpenetrating a pplied fi eld. On e of the two distinct tensor components is derived from res ults alread y obtained in co nnection wi th the electric polarizabili ty of short conducting cylinders. The other is calcu­ lated to an accuracy of four to five significan t fi gures for cylinders with radius to half-length ratios of }~, }~, 1, 2, a nd 4. These results, when combin ed with t he corresponding results for the electric polari zability, are a pplicable to t he proble m of calculating scattering from cylin­ cl ers and to the des ign of artifi c;al dispersive media. 1. Introduction This ar ticle considers the problem of finding the mftgnetie polariz ability tensor {3ij of a short, that is, noninfinite in length, righ t circular conducting cylinder under the assumption of negli gible field penetration. The latter condition can be realized easily with avail able con­ ductin g materials and fi elds which, although time varying, have wavelengths long compared with cylinder dimensions and can therefore be regftrded as quasi-static. T he approach of the present ar ticle is parallel to that employed earlier [1] 3 by the author fo r fin ding the electric polarizability of shor t co nductin g cylinders.
    [Show full text]
  • PH585: Magnetic Dipoles and So Forth
    PH585: Magnetic dipoles and so forth 1 Magnetic Moments Magnetic moments ~µ are analogous to dipole moments ~p in electrostatics. There are two sorts of magnetic dipoles we will consider: a dipole consisting of two magnetic charges p separated by a distance d (a true dipole), and a current loop of area A (an approximate dipole). N p d -p I S Figure 1: (left) A magnetic dipole, consisting of two magnetic charges p separated by a distance d. The dipole moment is |~µ| = pd. (right) An approximate magnetic dipole, consisting of a loop with current I and area A. The dipole moment is |~µ| = IA. In the case of two separated magnetic charges, the dipole moment is trivially calculated by comparison with an electric dipole – substitute ~µ for ~p and p for q.i In the case of the current loop, a bit more work is required to find the moment. We will come to this shortly, for now we just quote the result ~µ=IA nˆ, where nˆ is a unit vector normal to the surface of the loop. 2 An Electrostatics Refresher In order to appreciate the magnetic dipole, we should remind ourselves first how one arrives at the field for an electric dipole. Recall Maxwell’s first equation (in the absence of polarization density): ρ ∇~ · E~ = (1) r0 If we assume that the fields are static, we can also write: ∂B~ ∇~ × E~ = − (2) ∂t This means, as you discovered in your last homework, that E~ can be written as the gradient of a scalar function ϕ, the electric potential: E~ = −∇~ ϕ (3) This lets us rewrite Eq.
    [Show full text]
  • Ph501 Electrodynamics Problem Set 8 Kirk T
    Princeton University Ph501 Electrodynamics Problem Set 8 Kirk T. McDonald (2001) [email protected] http://physics.princeton.edu/~mcdonald/examples/ Princeton University 2001 Ph501 Set 8, Problem 1 1 1. Wire with a Linearly Rising Current A neutral wire along the z-axis carries current I that varies with time t according to ⎧ ⎪ ⎨ 0(t ≤ 0), I t ( )=⎪ (1) ⎩ αt (t>0),αis a constant. Deduce the time-dependence of the electric and magnetic fields, E and B,observedat apoint(r, θ =0,z = 0) in a cylindrical coordinate system about the wire. Use your expressions to discuss the fields in the two limiting cases that ct r and ct = r + , where c is the speed of light and r. The related, but more intricate case of a solenoid with a linearly rising current is considered in http://physics.princeton.edu/~mcdonald/examples/solenoid.pdf Princeton University 2001 Ph501 Set 8, Problem 2 2 2. Harmonic Multipole Expansion A common alternative to the multipole expansion of electromagnetic radiation given in the Notes assumes from the beginning that the motion of the charges is oscillatory with angular frequency ω. However, we still use the essence of the Hertz method wherein the current density is related to the time derivative of a polarization:1 J = p˙ . (2) The radiation fields will be deduced from the retarded vector potential, 1 [J] d 1 [p˙ ] d , A = c r Vol = c r Vol (3) which is a solution of the (Lorenz gauge) wave equation 1 ∂2A 4π ∇2A − = − J. (4) c2 ∂t2 c Suppose that the Hertz polarization vector p has oscillatory time dependence, −iωt p(x,t)=pω(x)e .
    [Show full text]
  • The Lorentz Force
    CLASSICAL CONCEPT REVIEW 14 The Lorentz Force We can find empirically that a particle with mass m and electric charge q in an elec- tric field E experiences a force FE given by FE = q E LF-1 It is apparent from Equation LF-1 that, if q is a positive charge (e.g., a proton), FE is parallel to, that is, in the direction of E and if q is a negative charge (e.g., an electron), FE is antiparallel to, that is, opposite to the direction of E (see Figure LF-1). A posi- tive charge moving parallel to E or a negative charge moving antiparallel to E is, in the absence of other forces of significance, accelerated according to Newton’s second law: q F q E m a a E LF-2 E = = 1 = m Equation LF-2 is, of course, not relativistically correct. The relativistically correct force is given by d g mu u2 -3 2 du u2 -3 2 FE = q E = = m 1 - = m 1 - a LF-3 dt c2 > dt c2 > 1 2 a b a b 3 Classically, for example, suppose a proton initially moving at v0 = 10 m s enters a region of uniform electric field of magnitude E = 500 V m antiparallel to the direction of E (see Figure LF-2a). How far does it travel before coming (instanta> - neously) to rest? From Equation LF-2 the acceleration slowing the proton> is q 1.60 * 10-19 C 500 V m a = - E = - = -4.79 * 1010 m s2 m 1.67 * 10-27 kg 1 2 1 > 2 E > The distance Dx traveled by the proton until it comes to rest with vf 0 is given by FE • –q +q • FE 2 2 3 2 vf - v0 0 - 10 m s Dx = = 2a 2 4.79 1010 m s2 - 1* > 2 1 > 2 Dx 1.04 10-5 m 1.04 10-3 cm Ϸ 0.01 mm = * = * LF-1 A positively charged particle in an electric field experiences a If the same proton is injected into the field perpendicular to E (or at some angle force in the direction of the field.
    [Show full text]
  • Section 14: Dielectric Properties of Insulators
    Physics 927 E.Y.Tsymbal Section 14: Dielectric properties of insulators The central quantity in the physics of dielectrics is the polarization of the material P. The polarization P is defined as the dipole moment p per unit volume. The dipole moment of a system of charges is given by = p qiri (1) i where ri is the position vector of charge qi. The value of the sum is independent of the choice of the origin of system, provided that the system in neutral. The simplest case of an electric dipole is a system consisting of a positive and negative charge, so that the dipole moment is equal to qa, where a is a vector connecting the two charges (from negative to positive). The electric field produces by the dipole moment at distances much larger than a is given by (in CGS units) 3(p⋅r)r − r 2p E(r) = (2) r5 According to electrostatics the electric field E is related to a scalar potential as follows E = −∇ϕ , (3) which gives the potential p ⋅r ϕ(r) = (4) r3 When solving electrostatics problem in many cases it is more convenient to work with the potential rather than the field. A dielectric acquires a polarization due to an applied electric field. This polarization is the consequence of redistribution of charges inside the dielectric. From the macroscopic point of view the dielectric can be considered as a material with no net charges in the interior of the material and induced negative and positive charges on the left and right surfaces of the dielectric.
    [Show full text]
  • Electromagnetic Induction, Ac Circuits, and Electrical Technologies 813
    CHAPTER 23 | ELECTROMAGNETIC INDUCTION, AC CIRCUITS, AND ELECTRICAL TECHNOLOGIES 813 23 ELECTROMAGNETIC INDUCTION, AC CIRCUITS, AND ELECTRICAL TECHNOLOGIES Figure 23.1 This wind turbine in the Thames Estuary in the UK is an example of induction at work. Wind pushes the blades of the turbine, spinning a shaft attached to magnets. The magnets spin around a conductive coil, inducing an electric current in the coil, and eventually feeding the electrical grid. (credit: phault, Flickr) 814 CHAPTER 23 | ELECTROMAGNETIC INDUCTION, AC CIRCUITS, AND ELECTRICAL TECHNOLOGIES Learning Objectives 23.1. Induced Emf and Magnetic Flux • Calculate the flux of a uniform magnetic field through a loop of arbitrary orientation. • Describe methods to produce an electromotive force (emf) with a magnetic field or magnet and a loop of wire. 23.2. Faraday’s Law of Induction: Lenz’s Law • Calculate emf, current, and magnetic fields using Faraday’s Law. • Explain the physical results of Lenz’s Law 23.3. Motional Emf • Calculate emf, force, magnetic field, and work due to the motion of an object in a magnetic field. 23.4. Eddy Currents and Magnetic Damping • Explain the magnitude and direction of an induced eddy current, and the effect this will have on the object it is induced in. • Describe several applications of magnetic damping. 23.5. Electric Generators • Calculate the emf induced in a generator. • Calculate the peak emf which can be induced in a particular generator system. 23.6. Back Emf • Explain what back emf is and how it is induced. 23.7. Transformers • Explain how a transformer works. • Calculate voltage, current, and/or number of turns given the other quantities.
    [Show full text]
  • Chapter 22 Magnetism
    Chapter 22 Magnetism 22.1 The Magnetic Field 22.2 The Magnetic Force on Moving Charges 22.3 The Motion of Charged particles in a Magnetic Field 22.4 The Magnetic Force Exerted on a Current- Carrying Wire 22.5 Loops of Current and Magnetic Torque 22.6 Electric Current, Magnetic Fields, and Ampere’s Law Magnetism – Is this a new force? Bar magnets (compass needle) align themselves in a north-south direction. Poles: Unlike poles attract, like poles repel Magnet has NO effect on an electroscope and is not influenced by gravity Magnets attract only some objects (iron, nickel etc) No magnets ever repel non magnets Magnets have no effect on things like copper or brass Cut a bar magnet-you get two smaller magnets (no magnetic monopoles) Earth is like a huge bar magnet Figure 22–1 The force between two bar magnets (a) Opposite poles attract each other. (b) The force between like poles is repulsive. Figure 22–2 Magnets always have two poles When a bar magnet is broken in half two new poles appear. Each half has both a north pole and a south pole, just like any other bar magnet. Figure 22–4 Magnetic field lines for a bar magnet The field lines are closely spaced near the poles, where the magnetic field B is most intense. In addition, the lines form closed loops that leave at the north pole of the magnet and enter at the south pole. Magnetic Field Lines If a compass is placed in a magnetic field the needle lines up with the field.
    [Show full text]
  • 1 Introduction 2 Spins As Quantized Magnetic Moments
    Spin 1 Introduction For the past few weeks you have been learning about the mathematical and algorithmic background of quantum computation. Over the course of the next couple of lectures, we'll discuss the physics of making measurements and performing qubit operations in an actual system. In particular we consider nuclear magnetic resonance (NMR). Before we get there, though, we'll discuss a very famous experiment by Stern and Gerlach. 2 Spins as quantized magnetic moments The quantum two level system is, in many ways, the simplest quantum system that displays interesting behavior (this is a very subjective statement!). Before diving into the physics of two-level systems, a little on the history of the canonical example: electron spin. In 1921, Stern proposed an experiment to distinguish between Larmor's classical theory of the atom and Sommerfeld's quantum theory. Each theory predicted that the atom should have a magnetic moment (i.e., it should act like a small bar magnet). However, Larmor predicted that this magnetic moment could be oriented along any direction in space, while Sommerfeld (with help from Bohr) predicted that the orientation could only be in one of two directions (in this case, aligned or anti-aligned with a magnetic field). Stern's idea was to use the fact that magnetic moments experience a linear force when placed in a magnetic field gradient. To see this, not that the potential energy of a magnetic dipole in a magnetic field is given by: U = −~µ · B~ Here, ~µ is the vector indicating the magnitude and direction of the magnetic moment.
    [Show full text]
  • Spin the Evidence of Intrinsic Angular Momentum Or Spin and Its
    Spin The evidence of intrinsic angular momentum or spin and its associated magnetic moment came through experiments by Stern and Gerlach and works of Goudsmit and Uhlenbeck. The spin is called intrinsic since, unlike orbital angular momentum which is extrinsic, it is carried by point particle in addition to its orbital angular momentum and has nothing to do with motion in space. The magnetic moment ~µ of silver atom was measured in 1922 experiment by Stern and Gerlach and its projection µz in the direction of magnetic fieldz ^ B (through which the silver beam was passed) was found to be just −|~µj and +j~µj instead of continuously varying between these two as limits. Classically, magnetic moment is proportional to the angular momentum (12) and, assuming this proportionality to survive in quantum mechanics, quan- tization of magnetic moment leads to quantization of corresponding angular momentum S, which we called spin, q q µ = g S ) µ = g ~ m (57) z 2m z z 2m s as is the case with orbital angular momentum, where the ratio q=2m is called Bohr mag- neton, µb and g is known as Lande-g factor or gyromagnetic factor. The g-factor is 1 for orbital angular momentum and hence corresponding magnetic moment is l µb (where l is integer orbital angular momentum quantum number). A surprising feature here is that spin magnetic moment is also µb instead of µb=2 as one would naively expect. So it turns out g = 2 for spin { an effect that can only be understood if linearization of Schr¨odinger equation is attempted i.e.
    [Show full text]
  • Magnetic Fields
    Welcome Back to Physics 1308 Magnetic Fields Sir Joseph John Thomson 18 December 1856 – 30 August 1940 Physics 1308: General Physics II - Professor Jodi Cooley Announcements • Assignments for Tuesday, October 30th: - Reading: Chapter 29.1 - 29.3 - Watch Videos: - https://youtu.be/5Dyfr9QQOkE — Lecture 17 - The Biot-Savart Law - https://youtu.be/0hDdcXrrn94 — Lecture 17 - The Solenoid • Homework 9 Assigned - due before class on Tuesday, October 30th. Physics 1308: General Physics II - Professor Jodi Cooley Physics 1308: General Physics II - Professor Jodi Cooley Review Question 1 Consider the two rectangular areas shown with a point P located at the midpoint between the two areas. The rectangular area on the left contains a bar magnet with the south pole near point P. The rectangle on the right is initially empty. How will the magnetic field at P change, if at all, when a second bar magnet is placed on the right rectangle with its north pole near point P? A) The direction of the magnetic field will not change, but its magnitude will decrease. B) The direction of the magnetic field will not change, but its magnitude will increase. C) The magnetic field at P will be zero tesla. D) The direction of the magnetic field will change and its magnitude will increase. E) The direction of the magnetic field will change and its magnitude will decrease. Physics 1308: General Physics II - Professor Jodi Cooley Review Question 2 An electron traveling due east in a region that contains only a magnetic field experiences a vertically downward force, toward the surface of the earth.
    [Show full text]